如何漂浮到熊猫中的对象？

Question

我有以下数据框： all_data

delay   settled_users   due_amt prime_tagging   pending_users   cycle_end_date
    0.0   114351    8.095711e+07    Prime_Super 236899           2022-03-15
    1.0   160691    5.590400e+07    Prime_Super 190559           2022-03-15
    2.0   211160    5.818422e+07    Prime_Super 140090           2022-03-15
    3.0   270745    7.271832e+07    Prime_Super 80505             2022-03-15

all_data.info()

<class 'pandas.core.frame.DataFrame'>
Int64Index: 30 entries, 1 to 6
Data columns (total 6 columns):
delay             30 non-null float64
settled_users     30 non-null int64
due_amt           30 non-null float64
prime_tagging     30 non-null object
pending_users     30 non-null int64
cycle_end_date    30 non-null object
dtypes: float64(2), int64(2), object(2)

我想要一个新列 all_data['delay'] + all_data['cycle_end_date'] 但这会抛出 TypeError: unsupported operand type (s) for +: 'int' 和 'datetime.date'

如何实现此目的？请帮忙

pd.to_timedelta(all_data['delay']) + pd.to_datetime(all_data['cycle_end_date'].astype(str))

或其他人 帮忙 结果

1   2022-03-15 00:00:00.000000000
2   2022-03-15 00:00:00.000000001
3   2022-03-15 00:00:00.000000002
4   2022-03-15 00:00:00.000000003
5   2022-03-15 00:00:00.000000004
6   2022-03-15 00:00:00.000000005
1   2022-03-15 00:00:00.000000000

Answer 1

使用 to_timedelta 用 unit='d' 表示天并将值添加到日期时间：

all_data['new'] = pd.to_timedelta(all_data['delay'], unit='d') + pd.to_datetime(all_data['cycle_end_date'])

或者：

all_data['new'] = pd.to_timedelta(all_data['delay'], unit='d') + pd.to_datetime(all_data['cycle_end_date'].astype(str))